Sub-tree with minimum color difference in a 2-coloured tree

A tree with N nodes and N-1 edges is given with 2 different colours for its nodes.
Find the sub-tree with minimum colour difference i.e. abs(1-colour nodes – 2-colour nodes) is minimum.

Examples:

```Input :
Edges : 1 2
1 3
2 4
3 5
Colours : 1 1 2 2 1 [1-based indexing where
index denotes the node]
Output : 2
Explanation : The sub-tree {1-2} and {1-2-3-5}
have color difference of 2. Sub-tree {1-2} has two
1-colour nodes and zero 2-colour nodes. So, color
difference is 2. Sub-tree {1-2-3-5} has three 1-colour
nodes and one 2-colour nodes. So color diff = 2.```

Method 1 : The problem can be solved by checking every possible sub-tree from every node of the tree. This will take exponential time as we will check for sub-trees from every node.

Method 2 : (Efficient) If we observe, we are solving a portion of the tree several times. This produces recurring sub-problems. We can use Dynamic Programming approach to get the minimum color difference in one traversal. To make things simpler, we can have color values as 1 and -1. Now, if we have a sub-tree with both colored nodes equal, our sum of colors will be 0. To get the minimum difference, we should have maximum negative sum or maximum positive sum.

• Case 1 When we need to have a sub-tree with maximum sum : We take a node if its value > 0, i.e. sum(parent) += max(0, sum(child))
• Case 2 When we need to have a sub-tree with minimum sum(or max negative sum) : We take a node if its value < 0, i.e. sum(parent) += min(0, sum(child))

To get the minimum sum, we can interchange the colors of nodes, i.e. -1 becomes 1 and vice-versa.

Below is the C++ implementation :

`// CPP code to find the sub-tree with minimum color`
`// difference in a 2-<a href="#">coloured tree</a>`
`#include <bits/stdc++.h>`
`using` `namespace` `std;`
`// Tree traversal to compute minimum difference`
`void` `dfs(``int` `node, ``int` `parent, vector<``int``> tree[], `
`                    ``int` `colour[], ``int` `answer[])`
`{`
`    ``// Initial min difference is the color of node`
`    ``answer[node] = colour[node];`
`    ``// Traversing its children`
`    ``for` `(``auto` `u : tree[node]) {`
`        ``// Not traversing the parent`
`        ``if` `(u == parent)`
`            ``continue``;`
`        ``dfs(u, node, tree, colour, answer);`
`        ``// If the child is adding positively to`
`        ``// difference, we include it in the answer`
`        ``// Otherwise, we leave the sub-tree and `
`        ``// include 0 (nothing) in the answer`
`        ``answer[node] += max(answer[u], 0);`
`    ``}`
`}`
`int` `maxDiff(vector<``int``> tree[], ``int` `colour[], ``int` `N)`
`{`
`       ``int` `answer[N + 1];`
`       ``memset``(answer, 0, ``sizeof``(answer));`
`    ``// DFS for colour difference : 1colour - 2colour`
`    ``dfs(1, 0, tree, colour, answer);`
`    ``// Minimum colour difference is maximum answer value`
`    ``int` `high = 0;`
`    ``for` `(``int` `i = 1; i <= N; i++) {`
`        ``high = max(high, answer[i]);`
`        ``// Clearing the current value`
`        ``// to check for colour2 as well`
`        ``answer[i] = 0;`
`    ``}`
`    ``// Interchanging the colours`
`    ``for` `(``int` `i = 1; i <= N; i++) {`
`        ``if` `(colour[i] == -1)`
`            ``colour[i] = 1;`
`        ``else`
`            ``colour[i] = -1;`
`    ``}`
`    ``// DFS for colour difference : 2colour - 1colour`
`    ``dfs(1, 0, tree, colour, answer);`
`    ``// Checking if colour2 makes the minimum colour `
`    ``// difference`
`    ``for` `(``int` `i = 1; i < N; i++)`
`        ``high = max(high, answer[i]);`
`        `
`    ``return` `high;`
`}`
`// Driver code`
`int` `main()`
`{`
`    ``// Nodes`
`    ``int` `N = 5;`
`    ``// Adjacency list representation`
`    ``vector<``int``> tree[N + 1];`
`    ``// Edges`
`    ``tree[1].push_back(2);`
`    ``tree[2].push_back(1);`
`    ``tree[1].push_back(3);`
`    ``tree[3].push_back(1);`
`    ``tree[2].push_back(4);`
`    ``tree[4].push_back(2);`
`    ``tree[3].push_back(5);`
`    ``tree[5].push_back(3);`
`    ``// Index represent the colour of that node`
`    ``// There is no Node 0, so we start from `
`    ``// index 1 to N`
`    ``int` `colour[] = { 0, 1, 1, -1, -1, 1 };`
`    ``// Printing the result`
`    ``cout << maxDiff(tree,  colour,  N);`
`    `
`    ``return` `0;`
`}`

Output:

```2

```

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source and credits.
source and credits: http://www.geeksforgeeks.org
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